We are tasked to solve the definite integral:

$\int_0^1 \frac{dx}{\sqrt{x} + 9\sqrt[3]{x}}$

Step 1: Simplify the denominator

The denominator is $\sqrt{x} + 9\sqrt[3]{x}$. Rewrite the terms using fractional exponents: $\sqrt{x} = x^{1/2}, \quad \sqrt[3]{x} = x^{1/3}.$ Thus, the denominator becomes: $x^{1/2} + 9x^{1/3}.$

Step 2: Substitution

Let $x = t^6$. Then $dx = 6t^5 dt$, and the powers of $x$ simplify: $x^{1/2} = (t^6)^{1/2} = t^3, \quad x^{1/3} = (t^6)^{1/3} = t^2.$ The integral bounds change as well:

• When $x = 0$, $t = 0$.
• When $x = 1$, $t = 1$.

The integral becomes: $\int_0^1 \frac{6t^5 \, dt}{t^3 + 9t^2}.$

Step 3: Simplify the integral

Factor the denominator: $t^3 + 9t^2 = t^2(t + 9).$ Thus, the integral simplifies to: $\int_0^1 \frac{6t^5}{t^2(t + 9)} \, dt = \int_0^1 \frac{6t^3}{t + 9} \, dt.$

Step 4: Further substitution

Let $u = t + 9$, so $du = dt$. When $t = 0$, $u = 9$. When $t = 1$, $u = 10$. Also, note $t = u - 9$, so: $t^3 = (u - 9)^3.$ The integral becomes: $\int_9^{10} \frac{6(u - 9)^3}{u} \, du.$

Step 5: Expand and integrate

Expand $(u - 9)^3$: $(u - 9)^3 = u^3 - 27u^2 + 243u - 729.$ Thus, the integral becomes: $\int_9^{10} \frac{6(u^3 - 27u^2 + 243u - 729)}{u} \, du = \int_9^{10} 6(u^2 - 27u + 243 - \frac{729}{u}) \, du.$

Separate into individual integrals: $6\int_9^{10} u^2 \, du - 162\int_9^{10} u \, du + 1458\int_9^{10} 1 \, du - 4374\int_9^{10} \frac{1}{u} \, du.$

Step 6: Compute each term

1. $\int u^2 \, du = \frac{u^3}{3}$: [ 6\left[\frac{u^3}{3}\right]_9^{10} = 2\left[10^3 - 9^3\right] = 2(1000 - 729) = 2(271) = 542. ]

2. $\int u \, du = \frac{u^2}{2}$: [ -162\left[\frac{u^2}{2}\right]_9^{10} = -81\left[10^2 - 9^2\right] = -81(100 - 81) = -81(19) = -1539. ]

3. $\int 1 \, du = u$: $1458\left[u\right]_9^{10} = 1458(10 - 9) = 1458.$

4. $\int \frac{1}{u} \, du = \ln|u|$: $-4374\left[\ln u\right]_9^{10} = -4374(\ln 10 - \ln 9) = -4374\ln\left(\frac{10}{9}\right).$

Final Answer:

Combine all terms: $\text{Integral} = 542 - 1539 + 1458 - 4374\ln\left(\frac{10}{9}\right).$

Let me know if you want more details or clarifications!

Related Questions:

1. How does substitution simplify integrals like this one?
2. Why is $\ln(x)$ used for integrating $\frac{1}{x}$?
3. How do definite integral bounds transform under substitution?
4. What methods can simplify the integration of rational functions?
5. How does the choice of substitution affect the difficulty of an integral?

Tip:

When dealing with roots and powers in integrals, consider substitutions that align with the powers (e.g., $x = t^n$). This often reduces complex expressions into simpler polynomial forms.